Base | Representation |
---|---|
bin | 101110010110011101… |
… | …111110001110111111 |
3 | 11202110111022012120202 |
4 | 232112131332032333 |
5 | 1303411341022433 |
6 | 34510312034115 |
7 | 3411216124565 |
oct | 562635761677 |
9 | 152414265522 |
10 | 49769079743 |
11 | 1a11a391269 |
12 | 978b69293b |
13 | 4901c7a916 |
14 | 25a1bbda35 |
15 | 146447dee8 |
hex | b9677e3bf |
49769079743 has 2 divisors, whose sum is σ = 49769079744. Its totient is φ = 49769079742.
The previous prime is 49769079721. The next prime is 49769079751. The reversal of 49769079743 is 34797096794.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 49769079743 - 220 = 49768031167 is a prime.
It is a super-3 number, since 3×497690797433 (a number of 33 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a self number, because there is not a number n which added to its sum of digits gives 49769079743.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (49769079443) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24884539871 + 24884539872.
It is an arithmetic number, because the mean of its divisors is an integer number (24884539872).
Almost surely, 249769079743 is an apocalyptic number.
49769079743 is a deficient number, since it is larger than the sum of its proper divisors (1).
49769079743 is an equidigital number, since it uses as much as digits as its factorization.
49769079743 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 72013536, while the sum is 65.
The spelling of 49769079743 in words is "forty-nine billion, seven hundred sixty-nine million, seventy-nine thousand, seven hundred forty-three".
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